External support reactions: RA= RC = RF = The shear and moment at point B are The moment at point C is KN KN KN kN-m. KN and kN-m, respectively.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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External support reactions:
RA =
RC =
RF =
The shear and moment at point B are
The moment at point C is
The moment at point D is
KN
KN
KN
kN-m.
kN-m.
The maximum positive moment at segment AC is
The maximum positive moment at segment CF is
The maximum shear along the entire span of the beam is
The maximum positive moment along the entire span of the beam is
The maximum negative moment along the entire span of the beam is
kN and
kN-m. It is located at
KN-m.
kN.
kN-m.
kN-m.
KN-m, respectively.
m from point A.
Express all your answers in three decimal places.
For the reactions, there is NO NEED to put the sign.
For V and M values, ALWAYS include the sign (+ or -) to denote the nature of shear and bending. Not indicating the sign will mark your answer wrong. No considerations.
Transcribed Image Text:External support reactions: RA = RC = RF = The shear and moment at point B are The moment at point C is The moment at point D is KN KN KN kN-m. kN-m. The maximum positive moment at segment AC is The maximum positive moment at segment CF is The maximum shear along the entire span of the beam is The maximum positive moment along the entire span of the beam is The maximum negative moment along the entire span of the beam is kN and kN-m. It is located at KN-m. kN. kN-m. kN-m. KN-m, respectively. m from point A. Express all your answers in three decimal places. For the reactions, there is NO NEED to put the sign. For V and M values, ALWAYS include the sign (+ or -) to denote the nature of shear and bending. Not indicating the sign will mark your answer wrong. No considerations.
Consider the following values in the given beam above:
L1=6m
L2 = 2 m
L3 = 4 m
L4=3m
L5=2 m
L6=1m
W1 = 90 kN/m
W2 = 30 kN/m
P = 50 kN
M = 60 kN-m
Point E is an internal hinge
W1
B
L2
L3
2 m
E
W2
L5
L6
H
Transcribed Image Text:Consider the following values in the given beam above: L1=6m L2 = 2 m L3 = 4 m L4=3m L5=2 m L6=1m W1 = 90 kN/m W2 = 30 kN/m P = 50 kN M = 60 kN-m Point E is an internal hinge W1 B L2 L3 2 m E W2 L5 L6 H
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